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BASIC program to calculate O/P of a heating panel

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Posted by Morris Dovey on January 7, 2007, 7:06 pm
 
I decided to take a shot at writing a simple program to predict the
output of a south-facing  vertical flat heating panel.

The starting point was the NREL web site; and I've collected the
applicable US maps on a web page at
http://www.iedu.com/DeSoto/SolarEnergy.html  - so that people can get a
quick kWh/m^2/day value to plug into the program. Be warned that there
are 12 monthly maps of the USA and that the page may take a while to
download.

According to what I've been able to find on the web, flat panels can
get an input boost from snow reflection that increases energy input
anywhere from 35 to 95 percent, depending on age of snow, surface
texture, etc. and the program allows entering a "snow boost"
percentage. People living in areas without snow should probably enter
a very small number to account for local reflective effects.

Panel area (in square feet) is, of course, an important input.

The program asks for a panel efficiency percentage - and this will
probably be a dealer-supplied number, a measured value, or a wild
guess (or perhaps a design target).

The program outputs a single value: the number of Btus that a panel
matching the supplied parameters can be expected to deliver.

I'm posting the code here to find out if I've made any glaring
errors - and to solicit refinements...

GW-BASIC code follows:

100 INPUT "Panel area in ft^2";AREA
110 INPUT "Panel efficiency (%)";EFFICIENCY
120 INPUT "Increased input from snow reflection (%)";SNOWBOOST
130 INPUT "Average daily sun in kWh/m2 from NREL map";KWHPERM2
140 BTUPERFT2 = KWHPERM2 * 316.998331#
150 BTUS = AREA * BTUPERFT2
160 BTUS = BTUS * (100 + SNOWBOOST) / 100
170 BTUS = BTUS * EFFICIENCY / 100
180 PRINT "Panel delivers";BTUS;"Btu per day"

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto



Posted by Duane C. Johnson on January 8, 2007, 3:01 am
 
Hi Morris;


 > I decided to take a shot at writing a simple
 > program to predict the output of a south-facing
 > vertical flat heating panel.

 > The starting point was the NREL web site; and I've
 > collected the applicable US maps on a web page at
 > http://www.iedu.com/DeSoto/SolarEnergy.html  - so
 > that people can get a quick kWh/m^2/day value to
 > plug into the program. Be warned that there are 12
 > monthly maps of the USA and that the page may take
 > a while to download.

 > According to what I've been able to find on the
 > web, flat panels can get an input boost from snow
 > reflection that increases energy input anywhere
 > from 35 to 95 percent, depending on age of snow,
 > surface texture, etc. and the program allows
 > entering a "snow boost" percentage. People living
 > in areas without snow should probably enter a very
 > small number to account for local reflective effects.

 > Panel area (in square feet) is, of course, an
 > important input.

 > The program asks for a panel efficiency percentage -
 > and this will probably be a dealer-supplied number,
 > a measured value, or a wild guess (or perhaps a
 > design target).

Or you could do a pretty good estimate using a
concept I call "Stasis Temperature". See a more
in depth explaination:
http://www.redrok.com/concept.htm#stasis

A fairly easy equation:
STASIS     = 340 Degrees Rankine at the 0%
                 efficiency temperature rise.
AMBT       = Ambient temperature of the panel.
OPPT       = Operating temperature of the panel.
EFFICIENCY = (STASIS + AMBT - OPPT) / STASIS * 100

 > The program outputs a single value: the number of
 > Btus that a panel matching the supplied parameters
 > can be expected to deliver.
 >
 > I'm posting the code here to find out if I've made
 > any glaring errors - and to solicit refinements...

Looks good to me.

 > GW-BASIC code follows:

 > 100 INPUT "Panel area in ft^2";AREA
 > 110 INPUT "Panel efficiency (%)";EFFICIENCY
 > 120 INPUT "Increased input from snow reflection (%)";SNOWBOOST
 > 130 INPUT "Average daily sun in kWh/m2 from NREL map";KWHPERM2
 > 140 BTUPERFT2 = KWHPERM2 * 316.998331#
 > 150 BTUS = AREA * BTUPERFT2
 > 160 BTUS = BTUS * (100 + SNOWBOOST) / 100
 > 170 BTUS = BTUS * EFFICIENCY / 100
 > 180 PRINT "Panel delivers";BTUS;"Btu per day"

10  CLS:COLOR 10
20  PRINT "Program to Estimate Solar Thermal"
30  PRINT "   Panel Performance Morris02.bas"
40  PRINT "      Written by Morris Dovey"
50  PRINT "         modified by Duane C. Johnson"
60  PRINT "            2006/01/07"
80  STASIS = 340 :REM Temperature rise of a flat plate
81               :REM  thermal panel with 2 layers of
82               :REM   transparent insulation running
83               :REM    at 0% efficiency.
100 COLOR 14
110 INPUT "                  Panel  area  in    ft^2 "; AREA
120 REM INPUT "                  Panel  efficiency   (%) "; EFFICIENCY
130 INPUT "                  Ambient   temperature F "; AMBT
140 INPUT "                  Operating temperature F "; OPPT
150 EFFICIENCY = (STASIS + AMBT - OPPT) / STASIS * 100
160 INPUT "Increased input from snow reflection  (%) ";SNOWBOOST
170 INPUT "Average daily sun in kWh/m2 from NREL map ";KWHPERM2
180 BTUPERFT2 = KWHPERM2 * 316.998331#
190 BTUS = AREA * BTUPERFT2
200 BTUS = BTUS * (100 + SNOWBOOST) / 100
210 BTUS = BTUS * EFFICIENCY / 100
220 COLOR 12
230 PRINT "                 Panel delivers Btu per day";BTUS
240 PRINT : GOTO 100

 > --
 > Morris Dovey
 > DeSoto Solar
 > DeSoto, Iowa USA
 > http://www.iedu.com/DeSoto

Duane

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Posted by Jeff on January 8, 2007, 6:22 am
 Duane C. Johnson wrote:

I was thinking along the same lines.

   I notice that the equation that you have on RedRock has a solar
concentration factor. It seems to me that CF is really a proportion to
what the solar insolation is. This would then be very close to the
instantaneous efficiency which is roughly:
Solar Insolation / T operating - T ambient

Certainly the efficiency under lower light is much less. That would
imply that the actual snow boost is greater than calculated.

Just my 2 cents.

   Jeff





Posted by nicksanspam on January 10, 2007, 1:28 pm
 

In BASIC :-)


It would be interesting to estimate the useful output of a solar heating
panel over a year from NREL's monthly average files for 239 stations. It
would be different for each month, and only useful if the average outdoor
temp were less than about 70 F.
 

... 95% is high. Page 238 of NREL's Blue Book says they use a ground
reflectance of 0.2. With a higher reflectance Rhog, we can add Iadj
= 0.5(Rhog-0.2)Ih(1-cos(beta)) to their published data, where Ih is
the global horizontal radiation and beta is the collector tilt.

Ih = 620 (Phila in Jan) and Beta = 90 degrees and Rhog = 0.6 (fresh snow)
makes Iadj = 0.5(0.6-0.2)620(1-0) = 124 Btu/ft^2 to add to the 1000 Btu/ft^2
of global sun on a south wall.


I tend to think of 4'x8' air heater boxes as expensive undersized toys,
producing the heat equivalent of about 1 gallon of oil per square foot
per year, a lot less than the annual heat requirement for an average house.
And they have sides and insulation board on the back, a lot more stuff than
an 8'x32' south wall covered with $/ft^2 R1 Dynaglas "solar siding" with
90% solar transmission.

Knowing the outdoor temp, we can estimate the heat gain directly.
If 1000 Btu/ft^2 falls on a south wall over 6 hours (say 2 hours less
than the calculated day length) on an average January day in Phila when
the outdoor temp is 34 F and the average collector air temp is 100 F,
the net heat gain is 0.9x1000-6h(100-34)1ft^2/R1 = 504 Btu/ft^2 per day.

Nick


Posted by nicksanspam on January 10, 2007, 2:19 pm
 

Oops. That was page 248.


And the number of daylight hours (equation 1.6.11 from my shiny new D&B)
is N = 2/15cos^-1(-tan(L)tan(D)), where L is the latitude and declination
D = 23.45sin(360(284+n)/365) and n is the day of the year.

Phila is about 40 n. lat, and D = -20.9 on an average January day,
so N = 2/15cos^-1(-tan(40)tan(-20.9)) = 9.5 hours, which agrees
with the nomogram on page 18. Maybe we should subtract 3 vs 2 hours
for the solar collection period.

Nick


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